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<br />TRIGONOMETRIC FORMUL/E
<br />e a e a c
<br />a
<br />Afb C �l � �. •
<br />Right Triangle Oblique_:l riangles ' +
<br />Solution of Right 9Triiiirglea
<br />T:..: b•
<br />For AnglerA -,sin= c ; cos = c , tan = b ;-cot =a sec R-..-
<br />.cosec
<br />Given ;Required : ` k _ ` (r° a
<br />a, b A; B ,c tan A = a = cot B', c x
<br />b :a2
<br />a c? Ei `YA B, b4 stn A",- ¢--
<br />t
<br />A, a } ! �$,: b, o B = 90°—A, b
<br />i
<br />A, b B a a B.=,90° Z i b tan A; a
<br />a a T4s '1; r -cos A
<br />B-'9 A,a ceinA,b caos.A c
<br />•._Solution of Oblrque`' Tnanglea}
<br />Givenequired' . 3 s `
<br />c lis; mei.B asin.CF
<br />,.0 180° (A -1- B) c R
<br />y c :, <sin:d.• `�* {. Y SET aln"A ''J
<br />t.�
<br />-gb
<br />A a b B, c; ;(!: sin'B r oe +.(A� B �� to sla C
<br />��t tr,aa ;(ab)tan'f1.k�)
<br />a b C A, B,j' Ai B 180 Qdtan`(��t
<br />z ,.r�,.,",, ,} a a ,�%r-:�, sln�% y�t' ay�,oF•t 1, r,• a'
<br />• ,.. _ r' , Vis, .� �4,�r�.: ,ti. a tr�`i 'r1P e'! �, ,.s�i"
<br />ab
<br />B, -a- s -stn? ( �'
<br />r t °B . C-180° _(A+B)
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<br />a, b, c `) r•�';.t¢+fib+cs,,"� �,`r"R`.;'�•.�j,,,�'�,:�-P*:;,�•�.
<br />*Area - c) _
<br />b,- e' 'Area`urea b o-siaaA�
<br />d, B, C;_a Area
<br />sa«;,,.:
<br />REDUCTION'.TQ HORIZONTAL
<br />Horizontal 'distance -Slope distance multiplied by the
<br />cosine ofthe.ve`rtical' le. Tbus: slope distance =319.4 ft.
<br />�spc 1. Vert. angle=5, 1 -From Tnhlo P.- nr ..,,a no vv—
<br />`-, + same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 50 101=.9959.1—.9959=.0041.
<br />319.4X:0041=1.31. 319.4-1.31=318.09 ft.
<br />l When the.rise is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise =14 ft.,
<br />slope distance=302.811 Horizontal distance=3026— 14 X 14 _3a6_0.32=302. 2S ft.
<br />2 X 3026
<br />NAGE IN V. S. A.
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